integral of x\sqrt[3]{x+1}

解答

\int x 3 x + 1 d x

\frac{3}{7} (x + 1)^{\frac{7}{3}} - \frac{3}{4} (x + 1)^{\frac{4}{3}} + C

求解步骤

\int x 3 x + 1 d x

使用换元积分法

= \int (u - 1) 3 u d u

乘开

(u - 1) 3 u : u^{\frac{4}{3}} - 3 u

= \int u^{\frac{4}{3}} - 3 u d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{\frac{4}{3}} d u - \int 3 u d u

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

\int 3 u d u = \frac{3}{4} u^{\frac{4}{3}}

= \frac{3}{7} u^{\frac{7}{3}} - \frac{3}{4} u^{\frac{4}{3}}

u = x + 1

代回

= \frac{3}{7} (x + 1)^{\frac{7}{3}} - \frac{3}{4} (x + 1)^{\frac{4}{3}}

解答补常数

= \frac{3}{7} (x + 1)^{\frac{7}{3}} - \frac{3}{4} (x + 1)^{\frac{4}{3}} + C

\int sin (7 x) d x

y^{^{''}} + 4 y = 4 t

d er i v a t i v e f (x) = (7^{- 3 x})

\int (4 x^{2} - \frac{2}{x}) d x

d er i v a t i v e f (x) = - x^{2} + 1